Another way to justify that $z+\dfrac1{z}=-1$ is that due to the coefficient symmetry, if $z$ is a root, then $\dfrac1{z}$ is a root as well. By Vieta, then, the sum of those two roots ought to be the negative of the coefficient of the linear term...
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Guess who it is.Jan 1 '12 at 15:15

Essentially the same calculation also follows from the observation that $x^2+x+1=\phi_3(x)$ is the third cyclotomic polynomial. So $z^3-1=(z-1)(z^2+z+1)=0$, and hence $z^3=1$ for any solution $z$. Therefore
$$
z^4+\frac{1}{z^4}=z\cdot z^3+\frac{(z^3)^2}{z^4}=z+z^2=-1.
$$

Very nice! If you had not posted this already, I would have used the sum of geometric series (instead of cyclotomic polynomials) to write $$0=z^2+z+1=\frac{z^3-1}{z-1}\Rightarrow z^3=1$$ and $$z^4+\frac{1}{z^4}=(z^3)z+\frac{1}{(z^3)z}=z+\frac{1}{z}=\frac{z^2+1}{z}=\frac{‌​-z}{z}=-1.$$
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Dilip SarwateJan 1 '12 at 15:48

Here is an alternative approach: let's consider $z^{8}+1$ , and then divide by $z^{4}$.

By using geometric series, notice that $$z^{8}+z^{7}+z^{6}+z^{5}+z^{4}+z^{3}+z^{2}+z+1=\left(z^{2}+z+1\right)\left(z^{6}+z^{3}+1\right)=0.$$ Now, as $z^{2}+z+1=0$, we know that both $z^{7}+z^{6}+z^{5}=0$ and $z^{3}+z^{2}+z=0$, and hence $$z^{8}+z^{4}+1=0$$ so that $z^{8}+1=-z^{4}$. Thus, we conclude that $$z^{4}+\frac{1}{z^{4}}=-1.$$

Different people see different things in the relation $z^2+z+1=0$. Just as Jyrki did, I see the third cyclotomic polynomial. Its roots are $-1/2 \pm \sqrt{-3}/2$, the two primitive cube roots of unity. Call one of these $\omega$ and see that $\omega^3=1$, so that $\omega^4=\omega$ and $\omega^{-4}=\omega^2$. Their sum is $-1$, from the defining relation.

$$z^2+z+1=0$$ Clearly $z \not= 0$ and therefore $$z+\frac{1}{z}=-1.$$
Note that $$z^{n+1}+\frac{1}{z^{n+1}}=(z+\frac{1}{z})(z^n+\frac{1}{z^n})-(z^{n-1}+\frac{1}{z^{n-1}}).$$
Therefore if we define the function $U:N \to N$ as $$U_n=z^n+\frac{1}{z^n}$$ then we have $U(0)=2$ and $U(1)=-1.$ Also $$U(n+1)=-U(n)-U(n-1)$$ for all $n \in N.$
Using this recurrence you can calculate $z^n+\frac{1}{z^n}$ for any $n \in N.$